mae243 section 1 fall 2006 class 2 darran.cairns@mail.wvu.edu

MAE243 Section 1 Fall 2006Class 2

darran.cairns@mail.wvu.edu

Three major design errors coupled with the underestimation of the dead load by 20% (estimated frame weight = 18 psf, actual frame weight = 23 psf) allowed the weight of the accumulated snow to collapse the roof. The load on the day of collapse was 66-73 psf, while the arena should have had a design capacity of at least 140 psf.

The top layer's exterior compression members on the east and the west faces were overloaded by 852%. The top layer's exterior compression members on the north and the south faces were overloaded by 213%. The top layer's interior compression members in the east-west direction were overloaded by 72%.

On January 18, 1978 the Hartford Arena experienced the largest snowstorm of its five-year life. At 4:15 A.M. with a loud crack the center of the arena's roof plummeted the 83-feet to the floor of the arena throwing the corners into the air. Just hours earlier the arena had been packed for a hockey game. Luckily it was empty by the time of the collapse, and no one was hurt

Snowstorm collapses roof hours after hockey game. No one hurt. Designers required to retake Mechanics of Materials!!

For those who can’t get enough of MAE 243

For the truly ambitious

FIG. 1-1 Structural members subjected to

axial loads

Resultant forces at O

To determine the load at any point in a material we consider it to be made up of many tiny cubes (finite elements). In general we assume that the material is continuous and cohesive. For each element we have three normal forces Fx, Fy and Fz. As we decrease the area the forces and area approach zero. However, the force/area normal to the area approaches a finite value. These are normal stresses (σ). Similarly, the force per unit area parallel to the surface are shear stresses (τ).

AFz

Az

0lim

AFx

Azx

0lim

A

FyAzy

0lim

General state of stress around the chosen point.

Units

SI N/m2 or Pa very often kPa, MPa & GPa are used.

US psi, kpsi

Assumptions

Bar is straight

Load is along centroid

Material is homogenous

Material is isotropic (or anisotropy is aligned with loading axis)

Localized distortions at the ends. Will not consider. If Bar is long compared to ends then don’t need to worry too much.

If bar is subject to constant uniform deformation then deformation is due to a constant normal stress

APAP

dAdfA

Normal Strain

sss

avg

sss

AB

lim

Tensile +ve

Compression -ve

FIG. 1-2Prismatic bar in tension:(a) free-body diagram of a segment of the bar, (b)

segment of the bar before loading, (c)

segment of the bar after loading, and (d) normal

stresses in the bar

A short post constructed from a hollow circular tube of aluminum supports a compressive load of 26 kips. The inner and outer diameter of the tube are 4.0 in and 4.5 in and its length is 16in. The shortening of the post due to the load is 0.0012 in.

Determine the compressive stress and strain in the post. (You may neglect the weight of the post and assume that buckling does not occur.)

The 80kg lamp is supported by two rods AB and BC as shown. If AB has a diameter of 10mm and BC has a diameter of 8mm, determine the average normal stress in each rod.

Get in groups and calculate the average normal stresses.

Internal Loading

X direction FBC(4/5)-FBAcos(60°)=0

Y direction

FBC(3/5)+FBAsin(60°)-80(9.81)=0

FBC=395.2 N

FBA=632.4 N

Average Normal Stress

MPam

N

A

F

MPam

N

A

F

BA

BABA

BC

BCBC

05.8)005.0(

4.632

86.7)004.0(

2.395

2

2

A circular steel rod of length L and diameter d hangs in a mine shaft and holds an ore bucket of weight W at its lower end.

(a) Obtain a formula for the maximum stress in the rod, taking into account the weight of the rod itself.

(b) Calculate the maximum stress if L=40 m, d=8mm and W=1.5 kN.

The slender rod is subjected to an increase of temperature along its axis, which creates a normal strain in the rod of

Where z is given in meters. Determine (a) the displacement of the end of the rod B due to the temperature increase and (b) the average normal strain in the rod.

2/13)10(40 zz

2/13)10(40 zz

dzzzd 2/13)10(401

Deformed length of small segment dz’

Integrate from 0 to 0.2 m to find total deformed length z’

mz

zzz

m

m

dzzz

20239.0

)3

2)(10(40

2.0

0

2/13

2.0

0

2/33

)10(401

Increase in length = 2.39mm

Average strain = 2.39/200=0.0119